Massa jenis amonium sulfat . Jika 264 g amonium sulfat dilarutkan dalam air hingga volume larutannya menjadi 2 L, hitung: Molaritas (M) Molalitas (m)

Pertanyaan

Massa jenis amonium sulfat $begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 space adalah space 1.290 space kg forward slash m cubed end style$ . Jika 264 g amonium sulfat dilarutkan dalam air hingga volume larutannya menjadi 2 L, hitung:

Molaritas (M)
Molalitas (m)

$begin mathsize 14px style left parenthesis A subscript r space N equals 14 comma space S equals 32 comma space O equals 16 space dan space H equals 1 right parenthesis end style$

Pembahasan Soal:

1. Menghitung molaritas $Error converting from MathML to accessible text.$

$open square brackets open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 close square brackets equals fraction numerator massa space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 space over denominator M subscript r space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end fraction cross times fraction numerator 1 over denominator V space larutan end fraction open square brackets open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 close square brackets equals fraction numerator 264 space gram over denominator 132 space gram space mol to the power of negative sign 1 end exponent end fraction cross times fraction numerator 1 over denominator 2 space L end fraction open square brackets open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 close square brackets equals 1 space M$

Jadi molaritas $Error converting from MathML to accessible text.$ adalah 1 M.

2. Menghitung molalitas $Error converting from MathML to accessible text.$

Molalitas adalah jumlah dari mol zat terlarut tiap 1000 gram pelarut.

Menentukan massa pelarut

Massa jenis $begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 space adalah space 1.290 space kg forward slash m cubed end style$ dapat dikonversi ke satuan $gram space cm to the power of negative sign 3 end exponent$ menjadi $1 comma 29 space gram space cm to the power of negative sign 3 end exponent$ artinya dalam setiap mL larutan terdapat massa larutan sebanyak 1,29 gram. Volume larutan dalam soal adalah 2000 mL, maka didapatkan massa larutan dan massa pelarut yaitu sebagai berikut: $table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row cell massa space jenis space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end cell equals cell fraction numerator massa space larutan space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 over denominator V space larutan space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end fraction end cell row cell 1 comma 29 space gram space cm to the power of negative sign 3 end exponent end cell equals cell fraction numerator massa space larutan space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 over denominator 2000 space cm cubed end fraction end cell row cell massa space larutan space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end cell equals cell 1 comma 29 space gram space cm to the power of negative sign 3 end exponent cross times 2000 space cm cubed end cell row cell massa space larutan space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end cell equals cell 2.580 space gram space end cell row blank blank blank row cell massa space larutan end cell equals cell massa space pelarut and massa space zat space terlarut end cell row cell massa space pelarut end cell equals cell massa space larutan bond massa space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end cell row cell massa space pelarut end cell equals cell 2580 space gram minus sign 264 space gram end cell row cell massa space pelarut end cell equals cell 2316 space gram end cell end table$

Menentukan molalitas $Error converting from MathML to accessible text.$

$table attributes columnalign right center left columnspacing 0px end attributes row cell m space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end cell equals cell fraction numerator massa space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 space over denominator M subscript r space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end fraction x fraction numerator 1000 over denominator massa space pelarut end fraction end cell row cell m space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end cell equals cell fraction numerator 264 space gram space over denominator 132 space gram space mol to the power of negative sign 1 end exponent end fraction x fraction numerator 1000 over denominator 2316 space gram end fraction end cell row cell m space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end cell equals cell 0 comma 86 space m end cell end table$

Jadi molalitas $Error converting from MathML to accessible text.$ adalah 0,86 m. $space$

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

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